Chemistry 125: Lecture 50 February 11, 2011 Electrophilic Addition with Nucleophilic Participation Cycloaddition Epoxides This For copyright notice see.

Chemistry 125: Lecture 50February 11, 2011

Electrophilic Addition withNucleophilic Participation

CycloadditionEpoxides This

For copyright notice see final page of this file

Problem:Suggest a Multi-Step Mechanism for the

Acid-Catalyzed “Pinacol Rearrangement”(draw nice curved arrows)

H+

CH3 C C

CH3

OH

CH3

CH3

OH

CH3 C C

CH3

CH3

CH3

O+ H2O

CH3 C C

CH3

+CH3

CH3

OH

CH3 C C

CH3

+CH3

CH3

O H

Methide Shift

+Driving Force?

Other “Simultaneous” ReagentsCl2C: (Carbene)

R2BH (Hydroboration)

CH2I2 Zn/Cu (Carbenoid)

O3 (Ozonolysis)

H-metal (Catalytic Hydrogenation)

R-metal (Metathesis, Polymerization)

RC (Epoxidation)OOH

O

Simmons-Smith“Carbenoid”

Metal R-X Metal+

R-X

single-electrontransfer(SET)

e

Metal+

R X Metal

R-M X +

Zn Cu“couple”

CH2I2 + CH2

The next three slides suggest a plausible, but incorrect, two-step mechanism for addition of ICH2ZnI to H2C=CH2

Cl Zn CH3

Model forI-Zn-CH2I

4sZn

LUMO 4pZnLUMO + 1

bent for transition stateLUMO` 4spnZn

HOMO

Model forI-Zn-CH2I

LUMO`

Zn-CHOMO

Model forI-Zn-CH2I

Cl Zn CH3

CH2 CH2

Cl Zn CH3I Zn CH2 I

HOMOZn-C

LUMO

C-I

ZnI2CH2

“SN2”

If it were the diiodide instead of the model…

But these two transition states were just guessed, not calculated quantum mechanically…

Although the above two-step mechanism with intermediate IZnCH2-CH2-CH2I is plausible,

addition of IZnCH2I to H2C=CH2 probably occurs in a single step,

according to quantum mechanical calculation*, with the bent

transition state shown below: * A DFT Study of the Simmons-Smith Cyclopropanation Reaction.

A. Bottoni, et al., J. Am. Chem. Soc, 1997, 119, 12300

ICH2ZnIZn

I

I

ICH2ZnI

(at TransitionState Geometry)

LUMO

Mixes with HOMO

HOMO-2

Mixes with * LUMO

Zn

I

I

meta-chlorobenzoic acid

meta-chloroperoxybenzoic acid

Epoxidation by Peroxycarboxylic Acids

J&F sec 10.4a 423-425

+ +

mCPBA

25°C

81% yieldR = n-hexyl

benzene5 hr

?

LUMO

HOMO-3

200

0

-200

-400

Orb

ital E

nerg

y (k

cal/m

ole)

UMOs

OMOs

etc.

Peroxyformic AcidDistorted to

Transition State for O Transfer

p(O

*O-O

“-allylic”

CH2

H2CCH2

H2C

O

OO C

H

H

“SN2 at O”

“-allylic” resonance

p()O

nucleophile(nearby)

*O-O electrophile

All happen together with

minimal atomic displacement

carboxylate “leaving group”

(but not strictly in parallel)

C-C nucleophile

pC+

electrophile

*H-O+

electrophile

“SN2 at H”backside

attack

Transition State

GeometryO-O

Strongly Stretched(from ~1.5Å)

O-HHardly

Stretched(from ~1Å)

kH/kD ~ 1

Coplanar“Butterfly”

mechanism(not spiro)

suggested by Paul D. Bartlett

(1950)

calculatedJ. Amer. Chem. Soc. (1991) pp. 2338-9

downhill motionafter TS

Only one TS :

“Concerted but not Synchronous”

“spiro” meanstwo perpendicular

rings sharing a common atom

(here O1)

Note that arrows were not used as carefully

in those days.

Bartlett 1950

Problem:How about now?

(compare arrows in this textbook illustration with the mechanism on

the previous frames and try drawing a more accurate diagram)

Stereospecificity of Epoxidation:Concerted Syn Addition

Pasto & Cumbo 1965

~0°C 10 hr

HH

C C

H3CCH3

O

>99.5% trans

mCPBAH

HC C

H3C

CH3

trans

O O

52-60% yield

O O

mCPBA

H HC C

H3C CH3

cis>99.5% cis

H H

C CH3C CH3

O

~0°C 10 hr 52-60% yield

Alternative Epoxide Preparation (1936)

H2O< 0°C3 hr

H HC C

H3C CH3 HOCl

Wilson & Lucas 1936

H H

C CH3C CH3

Cl+ H

H

C C

CH3

CH3

Cl

HO

55% yield(distilled)

SN2

H2O

H2O90°C2 hr

KOH(20M)

H

H

C C

CH3

CH3

Cl

-O

H HC CH3C CH3

O90% yield

45% over two steps

syn

inversion

2nd inversion

CH2CHO

HCC

H

OH

CH2

H

OR

ORO

OEtO

O CO2Et

TiRO

O

Remember Sharpless Asymmetric Epoxidation

R

ROO ••

RO+Ti

O

O

O

R

OEtO

CO2Et

O

RO

Ti

O

O

O OEtO

CO2Et

*

RCH2

HC

CCH

H

allyl alcohol

(R)-“epoxide”

(S)-epoxide precursor

Chiral“Oxidizing Agent”

LUMO?

HOMO?

is diastereomeric!

( also pO + *C=C )

Cf. J&F Sec. 10.4b p. 426

20,000,000 tons$20 billion

per year

OLDCAMPUS

H2C CH2

O H2C=CH2 + O2

Ag

250°C15 atm

ethylene oxide

(84%)

Raising the yield by 5% would be

worth >$109/year.

*

* The rest oxidizes

to CO2/H2O.

Only 0.05% of ethylene oxide is used as such.

H+ Catalysis

H2C CH2

O

20,000,000 tons$20 billion

per year

H2C CH2

O

ethylene glycol(antifreeze, solvents,

polymers)

J&F Sec. 10.4c pp. 427-430

H2C CH2

HO

OH

H2O

of which 2/3

H2C CH2

O

HO- Catalysis

End of Lecture 50February 11, 2011

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